\(\int \bigg( \sqrt x - \frac{1}{\sqrt x}\bigg)^2dx\)
= \(\int \bigg( x+\frac{1}{x}-2\bigg)dx\)
= \(\int xdx + \int \frac{1}{x} dx-2 \int 1.dx\)
= \(\frac{x^2}{2}+\log\mid x \mid -2x+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]

The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Integral calculus helps to resolve two major types of problems: